A hydrogen atom is in the n = 2 state. In the Bohr model, how many electron wavelengths fit around this orbit?
You have mutually exclusive concepts here. In the Bohr model, the electron is a particle with no wavelength.
At the quantum scale, the electron’s wave function spreads out over the larger volume of space it occupies at higher energies. It will become in essence a standing wave at the orbit, or if unable to stabilize, the excess energy is released as photons with well-known wavelength and the electron "drops" down into the ground state, and stabilizes as a "standing wave" in orbit.
I like Bekki’s analysis, but it assumes things that cannot be bourne out by experiment, just as the Bohr model does. The electron is a ½ integer spin fermion, and as a direct result, the wave functions will be standing half-waves.
Thus, the wavelength orbit will be ½n where n is the principal quantum number.
At n = 2, you will have one wavelength as the circumference.
Of course, all of the above ignores relativistic effects.
Bohr’s hypothesis was that for principal quantum number n, the angular momentum is n hbar.
From that, you can calculate that the orbital radius is n^2 times the bohr radius. Therefore the circumference also goes up like n^2.
So if the angular momentum goes up by a factor of n, and the radius goes up by n^2, the momentum must go down like 1/n.
The deBroglie wavelength goes like 1/p, so it is proportional to n.
So the number of wavelengths per circumference goes like n^2/n = n.
So if the ground state orbital is one wavelength, the answer for any other orbital would be n.
–And yes, what X says is correct. ANY attempt to analyze the Bohr model in detail means building upon assumptions that we now know are incorrect.